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Unformatted text preview: 1 Solution of Assignment No. ( 4 ) Kinematics of plane motion Intrinsic Coordinates 1 The velocity and the acceleration of a particle at point A of its path are indicated in the figure, where α=75ْ and β=45ْ. Determine: a The tangential and the normal component of acceleration at point A b The radius of the path at point A SOLUTION Part (a) The tangential acceleration represents the change in magnitude for the velocity w.r.t time 8 a t = a cosα = 100 cos(75 o45 o )= 100 cos(30 o )= 86.60 m/s 2 The normal acceleration represents the change in direction of the velocity 8 a n = a sinα = 100 cos(30 o )= 50 m/s 2 Part (b) The radius of curvature is given by: 8 a n = V A 2 /ρ 8 ρ = V A 2 / a n = (10) 2 /50 = 2m Part (c) The coordinates of the center of the path is C (x C , y C ) C A 45 o 75 o 30 o 45 o H a = 100m/s 2 v = 10m/s x a t a n O y t n 2 8 x C = x A – HA = 3 √ 2 – ρ sin(45 o ) = 2 √ 2 m 8 y C = y A + HC = √ 2 + ρ cos(45 o ) = 2 √ 2 m 2 The acceleration components of a particle motion are given by: a x = 40m/s 2 and a y = 30t m/s 2 Where t is in seconds. The initial conditions are: x o = 580m, y o = 0, s / m 25 y , x o . o . = = .When t = 1sec. Find the following: a The velocity and the acceleration of the particle b The time rate of change of the velocity magnitude c The coordinates of the center of curvature of the path at this instant SOLUTION ) 2 ( s / m t 30 a ) 1 ( s / m 40 a 2 y 2 x − − − = − − − = Initial conditions: at t = 0, x o = 580m, y o = 0, s / m 25 y , x o . o . = = Part (a) From equation 1 : ) 4 ( 580 t 20 x t 20 580 x tdt 40 dx t 40 dt dx x : And ) 3 ( t 40 x dt 40 x d 40 dt x d a 2 2 t x 580 . . t x . . x ....
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This note was uploaded on 12/21/2010 for the course ENGINEERIN mp108 taught by Professor Elbarki during the Spring '08 term at Alexandria University.
 Spring '08
 elbarki

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